PHYS207 Lecture Notes. J.K. Brosch Department of Physics and Astronomy, University of Delaware Sharp Lab, 102

PHYS207 Lecture Notes J.K. Brosch Department of Physics and Astronomy, University of Delaware Sharp Lab, 102 January 28, 2016

Contents 1 Measurement 3 2 Motion Along a Straight Line 4 2.1 Position, Displacement, and Average Velocity................. 4 2.2 Instantaneous Velocity.............................. 4 2.3 Acceleration.................................... 5 2.4 Constant Acceleration.............................. 5 2.5 Free-Fall Acceleration............................... 7 2.6 Graphical Integration in Motion Analysis.................... 7 3 Vectors 8 4 Motion in Two and Three Dimensions 11 4.1 Position and Displacement............................ 11 4.2 Average and Instantaneous Velocity....................... 11 4.3 Average and Instantaneous Acceleration.................... 12 4.4 Projectile Motion................................. 12 4.5 Uniform Circular Motion............................. 14 4.6 Relative Motion in One Dimension....................... 14 4.7 Relative Motion in Two Dimensions....................... 14 5 Force and Motion I 15 5.1 Newton s First and Second Laws........................ 15 5.2 Some Particular Forces.............................. 16 5.3 Applying Newton s Laws............................. 16 6 Force and Motion II 17 6.1 Friction...................................... 17 6.2 Drag Force and Terminal Speed......................... 17 6.3 Uniform Circular Motion............................. 17 7 Kinetic Energy and Work 19 7.1 Kinetic Energy.................................. 19 7.2 Work and Kinetic Energy............................ 19 7.3 Work Done by the Gravitational Force..................... 19 7.4 Work Done by a Spring Force.......................... 19 1

CONTENTS 2 7.5 Work Done by a Variable Force......................... 20 7.6 Power....................................... 21 8 Potential Energy and Conservation of Energy 22 9 Center of Mass and Linear Momentum 23 9.1 Center of Mass.................................. 23 9.2 Newton s Second Law for a System of Particles................ 23 9.3 Linear Momentum................................ 23 9.4 Collisions and Impulse.............................. 23 9.5 Conservation of Linear Momentum....................... 23 9.6 Momentum and Kinetic Energy in Collisions.................. 23 9.7 Elastic Collision in One Dimension....................... 23 9.8 Collisions in Two Dimensions.......................... 23 9.9 Systems with Varying Mass: A Rocket..................... 23 10 Rotation 28 11 Rolling, Torque, and Angular Momentum 29 11.1 Rolling as Translation and Rotation Combined................. 29 11.2 Forces and Kinetic Energy of Rolling...................... 30 11.3 The Yo-Yo..................................... 32 11.4 Torque Revisited................................. 33 11.5 Angular Momentum............................... 33 12 Equilibrium and Elasticity 34 13 Gravitation 35 14 Fluids 36 15 Oscillations 37

Chapter 1 Measurement We re going to skip this. 3

Chapter 2 Motion Along a Straight Line 2.1 Position, Displacement, and Average Velocity One of the primary things that we are interested in in physics, especially in classical mechanics, is the general position of a particle. Often time, this can be hard to find. It is often easier to know how fast something is going, or how much its speed is changing. One of the primary questions that we can ask is how much did a particle move. This leads us into our first definition. Definition 2.1.1. Suppose we have a particle that started at position x 1 and moves to position x 2. Then the displacement of the particle is given by x = x 2 x 1. (2.1.1) If now that we have a particle moving between to points, we might be interested in know how fast the particle was able to move between these two points. This leads us to the following definition. Definition 2.1.2. Suppose a particle starts at position x 1 at time t 1 and it moves position x 2 at a later time t 2. Then, the average velocity of the particle is 2.2 Instantaneous Velocity v avg = x t = x 2 x 1 t 1 t 2. (2.1.2) The problem with what we have done so far is that this only tells us what the average velocity of a particle is. But, what about if we want to know what happens to the velocity at any instant? The intuitive way of doing this would be to shrink our time interval. This concept gives us the next definition. Definition 2.2.1. The instantaneous velocity or simply velocity of a particle is given by the infinitesimal change in the displacement. i.e. x(t) x(t 0 ) v = lim = dx t t0 t t 0 dt. (2.2.1) Definition 2.2.2. We call the magnitude of the velocity of a particle the particle s speed. 4

CHAPTER 2. MOTION ALONG A STRAIGHT LINE 5 2.3 Acceleration Now that we have a notion of a particle s velocity, the next logical quantity to consider would be the change in a particle s velocity. We start, again, by first introducing the notion of change in velocity over a fixed time period. Definition 2.3.1. Suppose a particle initially has velocity v 1 at time t 1 and accelerates to velocity v 2 at a later time t 2. Then, the average acceleration of the particle is a avg = v t = v 2 v 1 t 1 t 2. (2.3.1) Again, this only gives us part of the picture. To truly understand acceleration, we will need to study the change in velocity over and infinitesimal time interval. This leads us to the next definition. Definition 2.3.2. The instantaneous acceleration or simply acceleration of a particle is given by the infinitesimal change in the average velocity. i.e. a = lim t t0 v(t) v(t 0 ) t t 0 = dv dt. (2.3.2) If we want to, we can relate the acceleration back to the position of a particle as follows: a = v = d dtẋ = ẍ. Here it is understood that the dot represents differentiation with respect to time. This is often called Newton s notation. I will primarily be using this notation going forward. It should be noted though, that the dot is reserved for only the time derivative. If we need to take a derivative with respect to another variable, we will use the standard notation. 2.4 Constant Acceleration Without making any further assumptions on the exact form the acceleration of a particle, it is often hard to make further progress in terms of saying much about the motion of a particle. Therefore, we will operate under the following assumption: the acceleration of the particle is constant. This simple assumption will take us a very long way. In fact, this assumption allows us to actually perform all of the physics that you will see in this book! To arrive at the basic equations of motion, we will need to perform some integration. First we start by rewriting (2.3.2) as dv = a dt. We will next integrate both sides of this. There are two ways we could proceed with this:

CHAPTER 2. MOTION ALONG A STRAIGHT LINE 6 1. specify initial conditions and use these equations to figure out what the constants of integration would be, 2. or, we could provide bounds for our integration. For those that have not had as much experience solving these types of equations, the first option is probably the most straightforward. On the other hand, being able to specify the right bound for the second method normally just takes experience before you know what to do. We will elect to go with the second method for easy and simplicity. For all of these derivations, we will assume that we can set our reference time such that t 1 = 0. For everything that appears in this book, this is a valid assumption. In fact, not only for this book, but for most 1 physics as a whole. We now return to what our original goal was, and integrate both sides of the rewritten (2.3.2), v t dv = a dt. v 0 0 Evaluating these integrals and rearranging gives us, For the next equation of motion, we rewrite (2.2.1) and integrate Substituting in (2.4.1) for v, we have x v = v 0 + at. (2.4.1) x 0 dx = t 0 v dt. Evaluating this integral gives us x x 0 = t 0 v 0 + at dt. x = x 0 + v 0 t + 1 2 at2. (2.4.2) Now that we have done this, we can solve (2.4.1) for the time t to get t = v v 0. a If we then substitute this expression for t into (2.4.2) and simplify, we obtain v 2 = v 2 0 + 2a(x x 0 ). (2.4.3) Equations (2.4.1) - (2.4.3) are the three primary equations that you will need for the course. An additional two equations may be derived by manipulating the first two of these three equations. 1 Generally any process that is symmetric under time reversal, or which has PT or PCT symmetry. A notable exception to this statement would be entropy.

CHAPTER 2. MOTION ALONG A STRAIGHT LINE 7 2.5 Free-Fall Acceleration If the movement of our particle is restricted entirely to vertical motion, and the only acceleration that the particle is subject to is gravity, then we say that the particle experience free-fall acceleration. We normally take the acceleration due to gravity to be g. 2.6 Graphical Integration in Motion Analysis Skipped

Chapter 3 Vectors So far, we have only talked about motion along a single line. But, as we know, particles are able to move in a large variety of exotic manners. To talk about this, we need to develop some tools that will serve us well for the rest of the course. We are, of course, talking about vectors. Before we define what a vector is, we should first define what a vector space is. Definition 3.0.1. Suppose we have a field K. In general, this can be over any field, either exotic or finite, but for our interests, R will be more than enough. We say that the set V is a vector space over the field K, if for all elements x, y, z in V and all scalars r, s in the field K the following eight properties hold: 1. Commutativity x + y = y + x 2. Associativity of vector addition (x + y) + z = x + (y + z) 3. Additive identity 0 + x = x 4. Additive inverse x + ( x) = 0 5. Associativity of scalar multiplication r(sx) = (rs)x 6. Scalar distribution (r + s)x = rx + sx 7. Vector distribution r(x + y) = rx + ry 8. Scalar identity 1x = x These eight rules give us everything we need now to define a vector. Definition 3.0.2. Given a vector space V, we say a vector is an element v of the vector space V. This is perhaps a bit more abstract definition of a vector than is given in most introductory physics books, but I feel that it being more explicit will be more beneficial to you. As a note on notation, must books will use bold face letters v to represent vectors, while in writing must people will either use an arrow above the letter v or a solid line under the letter v. Often time we can talk about a vector s components. Suppose we have a vector v with 8

CHAPTER 3. VECTORS 9 n components, then we can write v as v 1 v 2 v =., v n where each v i is a component of the vector v. If we want to add two vectors together, then we can simply add their components, i.e. if x 1 x 2 x =. x n y 1 y 2 y =., and we want to find x + y = z, then x 1 + y 1 x 2 + y 2 z = x + y =.. x n + y n Now suppose that we have a vector v that is just in R 2. If we place this vector such that it points radially out from the origin, then we can find v s x and y components. Suppose that from the positive x axis, the vector v goes out radially at an angle θ. Then, the x and y components are given by v x = v cos θ v y = v sin θ. Now that we have thoroughly talked about vector components, we can talk about a way of comparing two different vectors, namely comparing two vectors length. Definition 3.0.3. Let X be a vector space (over R or C), and let : X [0, ) have the following properties N1. x = 0 if and only if x = 0 N2. λx = λ x for all x in X and for all λ in K N3. x + y x + y for all x, y in X. Then we say that (X, ) is a normed space, and a norm. For the vectors we are concerned with in this course, the norm of a vector is going to be given by ( n ) 1 2 v 2 =. k=1 v 2 i y n

CHAPTER 3. VECTORS 10 This is often called the 2-norm of a vector, hence the 2 as the subscript. For the rest of this course, when we talk about the norm of the vector, we will understand that we will always mean the 2-norm even though the subscript will be left off. So far, we have only talk about vectors that have a notion of length, but we can go further than this and talk about vectors, with the notion of an angle between them. This leads us to the following notion, Definition 3.0.4. Let X be a vector space, and let (, ) : X X K have the following properties IP1. (x, y) = (x, y) IP2. IP3. (αx + βy, z) = α(x, z) + β(y, z) for all α, β in K, and for all x, y, z in X (x, αy + βz) = α(x, y) + β(z, z) for all α, β in K, and for all x, y, z in X IP4. (x, x) 0 and (x, x) = 0 if and only if x = 0. Then (, ) is an inner product, and ( X, (, ) ) is an inner product space. The inner product is also called a scalar product or dot product. In term of notation, the inner product between x and y will normally be written as (x, y), or x y. If these vectors are in R n, then we can apply the law of cosines to have where θ is the angle between them. x y = x y cos θ,

Chapter 4 Motion in Two and Three Dimensions 4.1 Position and Displacement Definition 4.1.1. The position of a particle as a function of time is given by r(t) = x(t)ˆx + y(t)ŷ + z(t)ẑ. (4.1.1) Definition 4.1.2. Suppose we have a particle that started at position r 1 and moves to position r 2. Then the displacement of the particle is given by r = r 2 r 1. (4.1.2) 4.2 Average and Instantaneous Velocity Definition 4.2.1. Suppose a particle starts at position r 1 at time t 1 and it moves position r 2 at a later time t 2. Then, the average velocity of the particle is v avg = r t = r 2 r 1 t 1 t 2. (4.2.1) Definition 4.2.2. The instantaneous velocity or simply velocity of a particle is given by the infinitesimal change in the displacement. i.e. It should be remembered here that v = lim t t0 r(t) r(t 0 ) t t 0 dr dt = dx(t) dt ˆx + dy(t) dt = dr dt. (4.2.2) ŷ + dz(t) ẑ, dt that is, differentiating a vector is equivalent to differentiating the vector component by component. 11

CHAPTER 4. MOTION IN TWO AND THREE DIMENSIONS 12 4.3 Average and Instantaneous Acceleration Definition 4.3.1. Suppose a particle initially has velocity v 1 at time t 1 and accelerates to velocity v 2 at a later time t 2. Then, the average acceleration of the particle is a avg = v t = v 2 v 1 t 1 t 2. (4.3.1) Definition 4.3.2. The instantaneous acceleration or simply acceleration of a particle is given by the infinitesimal change in the average velocity. i.e. 4.4 Projectile Motion a = lim t t0 v(t) v(t 0 ) t t 0 = dv dt. (4.3.2) Now that we understand how position, velocity, and acceleration behave in multiple dimensions, we are able to look at some classical scenarios where these can be used. Perhaps the most common application of what we have learned is projectile motion. In projectile motion, a particle is launched into the air at some angle, and we will want to find certain properties of the system that might arise. Claim 4.4.1. In projectile motion, the vertical and horizontal motion is independent of one another. That is, motion in x and y does not affect each other. Why would this be true? We will see this a we go through the rest of this section. We start by supposing that a ball in fired into the air at an angle θ to the horizontal. If we want to analyze this, we will need to identify what accelerations are affecting the particle. In y, we can clearly see that there is a negative acceleration due to gravity, namely a y = g. In x, in the absence of any external forces such as drag, buoyancy, friction, etc. there shouldn t be any acceleration. Therefor, we have a x = 0. Now that have figure out what the accelerations are, we can use Newton s first equation of motion, (2.4.2) in vector form, to get r r 0 = v 0 t + 1 2 at2. We can further brake this up into components to have ( ) ( ) x x0 v (x) 0 = y y 0 v (y) t + 1 ( ) 0 t 2. 0 2 g Since time is a scalar, we can pull it into our vectors (this is a direct consequence of Definition 3.0.1). This gives us ( ) ( ) x x0 v (x) 0 t = + 1 ( ) 0 y y 0 2 gt 2. v (y) 0 t

CHAPTER 4. MOTION IN TWO AND THREE DIMENSIONS 13 When we do this, we notice that no term with an x appears in the y equation, and vice-versa. This tells us that the systems are decoupled. Essentially, the motion in one dimension does not affect the motion in the other dimension. This proves Claim 4.4.1. Also, going back to chapter 3, and using the fact that v x = v cos θ v y = v sin θ, we can finally write down our equations of motion. Now that we have shown that they are in fact independent, we can split these equations apart to get our equations of motion, in x x x 0 = v 0 t cos θ, (4.4.1) and in y y y 0 = v 0 t sin θ 1 2 gt2. (4.4.2) As it stands now, we have given a parametric expression for the position of the projectile. But, there are numerous reason where this might not be desirable. In fact, it can often be advantageous to express the trajectory as a function of x, i.e. y(x). To do this, we solve for t in (4.4.1) and substitute this into (4.4.2). Doing this gives us y = x tan θ gx 2 2 (v 0 cos θ) 2. (4.4.3) The other main quantities that are of interest to us are the maximum range, R, of the particle and its maximum height h. To find the angle that gives us the maximum range, first start with (4.4.2) to solve for the total time of flight. Following this, you substitute this into (4.4.1) to get R = 2v2 0 g sin θ cos θ. We now use the trig. identity 2 sin θ cos θ = sin 2θ to get R = v2 0 g sin 2θ. We now take the derivative with respect to θ and set it equal to 0 to find 0 = 2v2 0 g cos 2θ. Inspecting this, we see that this means that 0 = cos 2θ, which gives us the condition that R is maximal when θ = π 2. Similarly, to find the maximum height for an arbitrary angle θ, we start with (4.4.2). We differentiate this with respect to t and set it equal to zero to get 0 = v 0 sin θ gt,

CHAPTER 4. MOTION IN TWO AND THREE DIMENSIONS 14 Solving for t we have t = v 0 sin θ. g This is the time it takes the projectile to reach its maximum velocity. We now take this time and plug it back into (4.4.2) to find h = (v 0 sin θ) 2 g (v 0 sin θ) 2 2g = (v 0 sin θ) 2. (4.4.4) 2g 4.5 Uniform Circular Motion skipped 4.6 Relative Motion in One Dimension Omitted 4.7 Relative Motion in Two Dimensions Omitted

Chapter 5 Force and Motion I Now that we are here, we are finally able to talk about what this course is really about. Namely, Newtonian Mechanics. This is, generally speaking, the study of forces applied to objects that are not too small 1, or too massive 2, or moving too fast 3. 5.1 Newton s First and Second Laws Newton s laws all have to do with motion of object, or to be more specific, the change in an object motion. This leads us to our first definition, namely, what causes a change in motion. Definition 5.1.1. A force is any interaction that, when unopposed, will change the motion of an object. Definition 5.1.2 (Newton s First Law of Motion). An object will not experience any acceleration unless it is acted on by an external force. There are many other statements of Newton s first law, but I think this is perhaps one of the cleanest ways of stating it. Often times, this can be referred to as the Law of Inertia. Though, considering how we defined force, we should look at this as an immediate consequence. The problem with this though, is that it is not quantitative. We need some way of quantifying this. This leads us to Newton s second law. Definition 5.1.3 (Newton s Second Law of Motion). Given a particle of mass m accelerating at a rate of a, the net external force is equal to the product of the mass and acceleration, i.e. F ext = ma. (5.1.1) You should think of Newton s second law as just a means of quantifying the first law. They are the same! 1 Meaning something above the scale where quantum effects are seen. 2 Meaning something below a scale where the mass is able to bend space time in general relativity. 3 Meaning that relativistic contractions present in special relativity are negligible. 15

CHAPTER 5. FORCE AND MOTION I 16 5.2 Some Particular Forces Probably the force we are all most familiar with is gravity. I m sure everyone has heard the (most certainly false) story of Newton discovering gravity after an apple fell on his head. For the majority of this course, we will assume that gravity is constant. You will experience non-constant gravity in chapter 13. In addition to this, we always assume that the force of gravity F G is always pointing downwards. Specifically, Definition 5.2.1 (Force due to Gravity). Suppose we have a particle of mass m acted on by a constant gravitational force g. Then the gravitational force F G is F = mg. Now that we have a concept of the force due to gravity, we can now define weight. Definition 5.2.2. The weight W of a particle is equal to the magnitude of the gravitational force F G acting on the particle. A lot of time people like to use the terms mass and weight interchangeably. They cannot be used interchangeably! A the mass of a particle is measure of the amount of matter. The weight of a particle is measure of force of gravity acting on a particular mass. 5.3 Applying Newton s Laws Definition 5.3.1. We say that two particles are interacting whenever there is a force acting between the two particles. Definition 5.3.2 (Newton s Third Law). Suppose we have two particles A and B. If A exerts a force F on B, then B exerts a force F on A. Definition 5.3.3. Suppose an object is in contact with a surface. Then the surface exerts a force on the object, called the normal force, that is always orthogonal to the surface.

Chapter 6 Force and Motion II 6.1 Friction Definition 6.1.1. Suppose an object is in contact with a surface. We say that the object experiences a (kinetic) frictional force if there is a force acting on the object, which is parallel to the surface, that opposed the motion of the object. We say the object experiences a (static) frictional force if there is a force acting on the object to prevent the object from moving. Friction has some interesting properties that you should always remember. Suppose we have an object in contact with some surface. Then, 1. if the object does not move, then the static frictional force f s is equal to the product of the coefficient of static friction µ s and the normal force F N, i.e. f s = µ s F n., and (6.1.1) 2. if the object is sliding along the surface, then the kinetic frictional force f k is equal to the product of the coefficient of kinetic friction µ k and the normal force F N, i.e. f k = µ k F N. (6.1.2) 3. In addition to this, µ k < µ s. That is, essentially, it is harder for an object to start moving than to continue moving. 6.2 Drag Force and Terminal Speed Skipped 6.3 Uniform Circular Motion Definition 6.3.1. Whenever a particle moves along a circle or circular arc at a constant velocity v, we say that it is moving in uniform circular motion. 17

CHAPTER 6. FORCE AND MOTION II 18 Now, when you think of this, you might realize that the particle must experience some kind of acceleration to be able to actually move in this circular path. That brings us to our next definition. Definition 6.3.2. Suppose a particle is moving in uniform circular motion with a constant velocity v. Then, the particle experience an acceleration, called centripetal acceleration. This acceleration is given by where R is the radius of the circular arc. a = v2 R, (6.3.1) Definition 6.3.3. A particle moving in uniform circular motion experiences a centripetal force, always point towards the center of the circle, which causes the centripetal acceleration. This force is given by F = m v2 R. (6.3.2) Figure 6.1: Figure illustrating the directions that the velocity and force point in uniform circular motion. Credit: WikiCommons. From the figure we can see that the force and velocity are always orthogonal to one another.

Chapter 7 Kinetic Energy and Work 7.1 Kinetic Energy Definition 7.1.1. The kinetic energy of a particle is the energy that the particle posses due to its motion. Suppose we have a particle of mass m moving at a velocity v. Then, the kinetic energy of the particle is 7.2 Work and Kinetic Energy T = 1 2 mv2. (7.1.1) Definition 7.2.1. Suppose we have an object that has some initial energy. Work is the energy transferred to or from the object due to some force. If work is done by a constant force. Then the work is Note that work is a scalar, not a vector. W = F d. (7.2.1) 7.3 Work Done by the Gravitational Force The work done by gravity is where θ is measured from the vertical. 7.4 Work Done by a Spring Force W G = mgd cos θ, (7.3.1) Definition 7.4.1. (Hooke s Law) Suppose we have some spring with spring constant k. Then the spring force is the negative of the product of the spring constant k and the the displacement d, i.e. F s = kd. (7.4.1) 19

CHAPTER 7. KINETIC ENERGY AND WORK 20 The negative sign in (7.4.1) signals that F s is a restoring force. This means that it is a force that wants to always return the spring to its equilibrium or relaxed state. This makes sense physically: if you compress a spring it want to decompress; if you stretch a spring, it want to snap back to its original position. In one dimension it is easiest to talk about the work done by a spring. Theorem 7.4.1. Suppose we have a spring with spring constant k that starts at position x i and is stretched to x f. Then, the work done by the spring force is given by W s = 1 2 k(x2 i x 2 f). (7.4.2) Proof. Suppose we have a spring with spring constant k that starts at position x i and is stretched to x f. The work done by the spring will be W s = F s dx. Substituting in (7.4.1) in this and applying our bounds of integration we have W s = k We now reverse our bounds of integration to cancel the minus sign out front. Twe then integrate to get xi W s = k x dx = 1 x f 2 kx2 x i. x f Evaluating these limits gives us and we are done. xf x i x dx W s = 1 2 k(x2 i x 2 f), 7.5 Work Done by a Variable Force I have absolutely no idea why this doesn t come before all the other sections on work in this chapter, but we ll go with it... Definition 7.5.1. Suppose we have some variable force F. Then, if we want to calculate the work done by a variable force, we have W = F dl. (7.5.1)

CHAPTER 7. KINETIC ENERGY AND WORK 21 For what we are interested in, dl will be dl = dxˆx + dyŷ + dzẑ. (7.5.2) Substituting (7.5.2) into (7.5.1) we have W = F x dx + F y dy + F z dz. (7.5.3) Theorem 7.5.1 (Work-Kinetic Energy Theorem). The work done by a force F on a particle of mass m is equal to the change in kinetic energy. Proof. For simplicity, we will assume that this is only one dimension, i.e. F = F (x). The result hold for higher dimensions and follows a similar argument. Then, using the one dimension version of (7.5.3), we have xf xf W = F (x) dx = m a dx, x i x i where we have substituted in Newton s second law on the right hand side. We, again, rewrite the right hand side use the fact that a = dv. We now apply the chain rule for differentiation dt to dv to get dt dv dt = dv dx dx dt = dv dx v. Thus, we now have vf W = m v dv, v i where we have changed our limits of integration according to the chain rule. Evaluating this give us W = 1 v f 2 mv2 = 1 v i 2 mv2 f 1 2 mv2 i. We immediately recognize the two terms to be kinetic energies. Thus, we have and we are done. W = T f T i = T, 7.6 Power skipped

Chapter 8 Potential Energy and Conservation of Energy 22

Chapter 9 Center of Mass and Linear Momentum 9.1 Center of Mass 9.2 Newton s Second Law for a System of Particles 9.3 Linear Momentum 9.4 Collisions and Impulse 9.5 Conservation of Linear Momentum 9.6 Momentum and Kinetic Energy in Collisions 9.7 Elastic Collision in One Dimension 9.8 Collisions in Two Dimensions 9.9 Systems with Varying Mass: A Rocket So far in this chapter, we have assumed that the total mass of the system always remains constant. There are some scenarios were this is obviously not true. Probably the simplest one to imagine and solve is that of a rocket. Most of the initial mass at the time of launch in a rocket is made up of fuel which will be burned as the rocket moves. As you might already be able to see, if we consider the varying mass of the rocket itself, then the mass would obviously not be constant as the rocket accelerates. If instead of this, if we consider our system to be the rocket and the fuel that it ejects as it accelerates, then 23

CHAPTER 9. CENTER OF MASS AND LINEAR MOMENTUM 24 the mass of this entire system would remain constant. We now proceed to develop what is called the rocket equation use Newton s second law. Figure 9.1: Illustration of the rocket at time t (left) and the rocket again at time t + dt (right). For this problem, we will assume that the mass is burnt off of the rocket at a rate m(t). This burning off produces an exhaust velocity v ex, which we take to be constant. In the absence of an external force, (total) linear momentum is conserved. From Figure 9.1 we see that the momentum of the rocket at time t is simply p i = mv. At time t + dt, we can also look at Figure 9.1 again to see that the momentum of this is p f = (m + dm)(v + dv) + ( dm)(v v ex ) mv + mdv + vdm vdm + v ex dm = mv + mdv + v ex dm. In this process at time t + dt, you might have noticed that a dmdv term did not appear. You can think of why this doesn t appear in this way: first, note that both dm and dv are infinitesimal changes in m and v. As such, we understand that these quantities are both very small. Thus, the product of the two is even smaller, especially when compared to quantities such as mv and v ex dm. Therefore, we may say that its contribution is negligible. Therefore, applying conservation of momentum gives us mv = mv + mdv + v ex dm 0 = mdv + v ex dm dm m = dv v ex. If we assume that the rocket starts with a mass m 0 and velocity v 0, we can integrate both

CHAPTER 9. CENTER OF MASS AND LINEAR MOMENTUM 25 sides of this to m and v respectively to get Rearranging this gives us m m 0 dm m = v v 0 dv v ex ln m m = v v 0 m 0 v ( ) ex m ln = v v 0. v ex m 0 ( ) m(t) v = v 0 v ex ln, (9.9.1) which is often called the rocket equation. This is the point where your book leaves off on this, but we can make a bit more progress in terms of understanding the physics of what is happening with the rocket. In general, it is often hard to find the position of the rocket from this velocity equation because of m(t). For most functions of t, this won t be possible, so we will make a somewhat realistic assumption on the functional form, i.e. m 0 m(t) = m 0 kt, where k > 0. Note that is valid for 0 t m 0. This assumption basically means that we k expend fuel at a constant rate. Thus, to now get y(t), we integrate (9.9.1) once, t ( ) m(t ) t ( ) y y 0 = v 0 v ex ln dt m0 kt = v 0 t v ex ln dt 0 m 0 0 m 0 t ) = v 0 t v ex ln (1 km0 t dt 0 = v 0 t + v ex m 0 k = v 0 t + v ex m 0 k = v 0 t + v ex m 0 k [(1 km0 t ) ( 1 k ) t m 0 ) ( ln (1 km0 t 1 k )] t t m 0 0 ( ln 1 k ) ( t 1 k ) t m 0 m 0 [( 1 k ) ( t ln 1 k ) t + k ] t. m 0 m 0 m 0 ln(1) 1 }{{} =0 Therefore, when we consider linear fuel expendature, we see that the position y(t) of the rocket is ( ( ) ) m(t) m(t) y y 0 = v 0 t v ex k ln t. (9.9.2) Throughout this entire process, we have assumed that the rocket was operating without the influence of gravity. When we normally think of rockets, we are immediately drawn to images of the Saturn V rocket being launched to go to the moon. In this scenario, gravity m 0

CHAPTER 9. CENTER OF MASS AND LINEAR MOMENTUM 26 will clearly play a role. A rocket leaving an atmosphere is generally very difficult to derive an exact expression, mostly because gravity changes with distance from the surface of the planet, so you cannot treat it as constant. You will see exactly why when you get to chapter 13. If instead of this, we consider, say, a model rocket being launched in your backyard, then we can assume that gravity is constant, which leads us to the following claim. Claim 9.9.1. If a rocket is moving in a constant gravitational field, with a constant force of gravity g, then (9.9.1) becomes ( ) m(t) v = v 0 v ex ln gt, m 0 and (9.9.2) becomes y y 0 = v 0 t v ex ( m(t) k ( ) m(t) ln m 0 ) t gt2 2. So, why should this be true? If we go back to our original integral for the velocity equation, we will notice the adding on the constant term g doesn t really change much since it has no t dependence. Thus, when we integrate this term, which is linear in t, we would simply pick up a gt, and similarly for the y(t) equation. Now that we have finally concluded all the remarks that need to be made about the rocket problem, we will turn our attention to some sample problems to help reinforce what we have just done. Example 9.9.1. Consider a rocket of initial mass m 0 accelerating from rest in free space. At first, as it speeds up, its momentum p increases, but as its mass m decreases, p eventually begins to decrease. For what value of m is p maximal? Solution: To solve this, we start with the rocket equation (9.9.1) with v 0 = 0. The rocket s linear momentum is given by ( ) m p = mv = mv ex ln. m 0 We now differentiate p with respect to m to find when p is maximal. ( ) dp m dm = v 1 ex ln mv ex m 0 m ( ) ) m = v ex (ln + 1. Looking at this, we can see that dp dm = 0 when m m 0 = 1 e. m 0

CHAPTER 9. CENTER OF MASS AND LINEAR MOMENTUM 27 The next example has to deal with the difference between single stage and multistage rockets. Unlike with a model rocket, rockets like the Saturn V are multistage, which gives them more thrust. We will now investigate how much advantage this gives versus a single stage rocket. Example 9.9.2. Consider the following (a) A certain rocket carries 60% of its initial mass as fuel. That is, the mass of the fuel is 0.6m 0. What is the rockets final speed if it is accelerating from rest in free space, if it burns all of its fuel in a single stage? The answer should be expressed as a multiple of v ex. (b) Suppose instead of burning all of its fuel in one stage, we have a rocket that burns its fuel in two stages as follows: in the first stage, it burns a mass 0.3m 0 of fuel. It then jettisons the first stage fuel tank, which has a mass of 0.1m 0, and then burns the remaining 0.3m 0 of fuel. Find the final speed of the rocket, again expressing the answer in terms of v ex. How does this compare to the answer in the first part? Solution For this problem, we will use the rocket equation (9.9.1). To make things easier, we will rewrite this slightly as ( ) m0 v f = v ex ln. m(t) (a) For the single stage rocket, since we burn off 60% of the fuel we have that m = m 0 0.6m 0 = 0.4m 0. Thus ( ) m0 v f = v ex ln = v ex ln 2.5. 0.4m 0 (b) For the first stage, we burn off 0.3m 0 off mass thus, the mass at the end of stage one is 0.7m 0. Thus the velocity for the first stage is given by ( ) m0 v f = v ex ln = v ex ln 10 0.7m 0 7. We now jettison the first stage rocket, which has a mass 0.1m 0, to arrive at the initial mass at the start of the second stage is 0.6m 0. We then burn off 0.3m 0 of fuel to arrive at a final mass for stage two of 0.3m 0. Thus, the velocity of the second stage is v f = v 1 + v ex ln 0.6m 0 = v 1 + v ex ln 2 0.3m ( 0 = v ex ln 10 ) 7 + ln 2 = v ex ln 20 7 = v ex ln 2.85. Using the fact that ln 2.5 = 0.92 and ln 2.85 = 1.04, we can see that there is over a 10% increase in the final velocity using a multistage rocket. Obviously, the more stages you use, the more of an advantage you will get.

Chapter 10 Rotation 28

Chapter 11 Rolling, Torque, and Angular Momentum 11.1 Rolling as Translation and Rotation Combined Definition 11.1.1. We say an object rolls smoothly across a surface, if it rolls without slipping or bumping. In a sense, all of our interests will be in motion which we consider suitably nice. Studying rolling is perhaps one of the most difficult physical problems that we have considered so far. One simple reason this can be is because even though the center of the object moves in a straight line, any point on the rim of the object does not. The way to study this is to break the motion up into a combination of linear and rotational motion. In the previous chapter, we related the arc length s to the rotational angle θ: s = θr, where R is the radius of the ring. Note that for our wheel, the center of the wheel is also the center of mass. The linear speed of the center of the would be given by ds, while the dt angular speed of the wheel about its center would be dθ. dt Therefore, we differentiate both side of the equation above to get v com = ωr. (11.1.1) Looking at Figure 11.1, we see that rolling motion is the linear combination of pure rotation and pure translation. In pure rotation, every point on the wheel rotates about the center with an angular speed ω. In pure translation, all points on the wheel move with speed v = v com. To get rolling motion, we simply add the two different scenarios together vectorally. When we do this, we see the top of the wheel has velocity 2v and the bottom of the wheel has 29

CHAPTER 11. ROLLING, TORQUE, AND ANGULAR MOMENTUM 30 Figure 11.1: Figure of the rolling motion of a wheel. The left most wheel is pure rotation motion. The center wheel is pure linear motion. The right most wheel is the combination of both the left and center wheels to make rolling motion. velocity of 0. This is purely a consequence of us adding the two scenarios together vectorally. Up until now, we have viewed rolling as a combination of translation and rotation, but there is another way that we can view this. Namely, we can view rotation as pure rotation about and axis that is passing through the point P. 11.2 Forces and Kinetic Energy of Rolling Now that we have gone through the process of formulating the rolling problem, one of the natural things to ask about it is, what might it s kinetic energy be. One way of calculating this would be to view the problem again as just the rotation about the point P. Then, using the formula for kinetic energy from last chapter, we have T = 1 2 I pω 2. Now, this isn t the most useful expression. What we should do is use the parallel-axis theorem to get I for the center of the wheel, i.e. I p = I com + MR 2. We now substitute this into our expression for the kinetic energy and we have T = 1 2 I comω 2 + 1 2 MR2 ω 2. Using (11.1.1), we ca rewrite the second term as T = 1 2 I comω 2 + 1 2 Mv2 com. (11.2.1) When looking at this expression, we see that in fact, this is just a combination of the kinetic energies for translation and rotation, which how we originally formulated this problem.

CHAPTER 11. ROLLING, TORQUE, AND ANGULAR MOMENTUM 31 Similarly with how we proceeded back in chapter 5 and chapter 6, we will be interested in things like a ball rolling down a hill. In addition to this, we will also be interested in how friction affects rolling motion. We first start by differentiating (11.1.1) by t to get We now consider a wheel rolling down a ramp with friction. a com = αr. (11.2.2) Figure 11.2: Wheel rolling down the ramp To start, we first draw the diagram given by Figure 11.2. We then apply Newton s second law in angular and linear motion. Here are some notes to keep in mind. 1. The gravitational force F G on the body is directed downwards, The component of this along the ramp is F G sin θ which is equal to mg sin θ. 2. A normal force F N is perpendicular to the ramp. It acts on the point of contact P. 3. A static frictional force f s acts on the point P and is directed up the ramp. Therefore, along the x axis, Newton s second law becomes f s mg sin θ = ma (x) com. (11.2.3) We now look to apply Newton s second law in angular form. We do this for an axis about the body s center of mass, i.e. Rf s = I com α. (11.2.4)

CHAPTER 11. ROLLING, TORQUE, AND ANGULAR MOMENTUM 32 From our diagram, it can be found that α = a(x) com R. We substitute this into the above equation to get a (x) com f s = I com R. (11.2.5) 2 Substituting this into equation (11.2.3) yields a (x) com = g sin θ 1 + Icom mr 2. (11.2.6) When we do this, you should note that gravity is what causes the ball to come down the ramp, but it friction that causes the ball to rotate, and the roll. Example 11.2.1. Suppose a uniform ball of mass m and radius R rolls smoothly from rest down a ramp at an angle θ. If the ball descends a vertical height h, what is the speed of the ball at the bottom? Solution: To solve this we will use conservation of energy. Since it is at rest the initial kinetic energy is 0. Thus the only initial energy is the potential energy mgh. When the ball is at the bottom of the ramp, there is no more potential energy, so there is just the energy from rolling. Thus, we have Rearranging this gives 11.3 The Yo-Yo 1 2 I comω 2 + 1 2 mv2 com = mgh. 2mgh Icom v =. m Now, people could think of the yo-yo as being a totally different problem than the ball rolling down the ramp, but I claim that it is not. In fact, I claim that it is nothing more than one of the limiting cases of the ball rolling down the ramp. For a yo-yo, the motion is purely vertical, with the yo-yo rotating as it moves up and down the string. The analysis we did for the ramp is almost identical to this analysis except for a few things: 1. Instead of rolling down a ramp at an angle θ with the horizontal, the yo-yo rolls downs a string with an angle θ = π 2. 2. Instead of rolling on its outer surface at a radius of R, the yo-yo rolls on an axle of radius R 0. 3. Instead of being slowed by frictional force f s, the yo-yo is slowed by the force F T on the string. Therefore, going through the same identical process again, we see that a com = g. (11.3.1) 1 + Icom mr0 2

CHAPTER 11. ROLLING, TORQUE, AND ANGULAR MOMENTUM 33 11.4 Torque Revisited Figure 11.3: Figure of a yoyo. Definition 11.4.1. Let r be a position vector, and F a force acting in the plane of the particle. The torque τ is the result given by the cross product of position and force, i.e. τ = r F. (11.4.1) If we want to find the magnitude of the torque, we simply apply the law of sines to this to get τ = rf sin θ. (11.4.2) 11.5 Angular Momentum Definition 11.5.1. Consider a particle of mass m moving past a point in the xy plane. The angular momentum of this particle is given by where r is the position vector to that particular point. l = r p = m(r v), (11.5.1) To find the direction of the angular momentum, we use the standard right hand rule for vector cross products. If we wish to find the magnitude of the angular momentum, then again, we just apply the law of sines to this to get l = rmv sin θ. (11.5.2) It should be noted that angular momentum only has meaning with respect to a specified origin, and its direction is always orthogonal to the plane formed by the two vectors r and p.

Chapter 12 Equilibrium and Elasticity skipped 34

Chapter 13 Gravitation 35

Chapter 14 Fluids skipped 36

Chapter 15 Oscillations 37